How do you solve \[{{\left( x-5 \right)}^{2}}=4\]?
Answer
624k+ views
Hint: We can solve this question by basic linear equation concept. First we have to expand the LHS as quadratic equation and then we have to solve it. We will use factoring methods to solve the equation. We have to split the middle term and take common terms to find factors. From the factors we can find the values of \[x\].
Complete step by step answer:
Given equation
\[{{\left( x-5 \right)}^{2}}=4\]
First we have to expand the LHS using the formula
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
By applying above formula we can expand our LHS as
\[{{\left( x-5 \right)}^{2}}={{x}^{2}}-2\times x\times 5+{{5}^{2}}\]
From this we will get
\[\Rightarrow {{x}^{2}}-10x+25\]
Now we have to substitute this in our equation. Our equation will look like
\[\Rightarrow {{x}^{2}}-10x+25=4\]
Now we will solve this quadratic equation using the factor method.
For this we have to make our RHS as \[0\] by transferring \[4\] to LHS. We will get
\[\Rightarrow {{x}^{2}}-10x+21=0\]
For solving using factor methods we have to split the middle term.
We have split the middle term in such a way that their sum is equal to the middle term and product is equal to first and last terms.
To get these numbers we have to find prime factorization of the product of first and last terms.
The product we will get from our equation is \[21\]
The prime factorization of \[21\] will be like
\[\Rightarrow 21=3\times 7\]
So we can see that \[-3\] and \[-5\] can satisfy our condition.
\[\begin{align}
& -3\times -7=21 \\
& -3+\left( -7 \right)=-10 \\
\end{align}\]
So we can split our middle term now. By splitting it we will get
\[\Rightarrow {{x}^{2}}-3x-7x+21=0\]
Now we have to take common terms .
From the equation we can take \[x\] as common from the first two terms and \[7\] as common from the next two terms.
Then the equation will look like
\[\Rightarrow x\left( x-3 \right)-7\left( x-3 \right)\]
\[\Rightarrow \left( x-3 \right)\left( x-7 \right)\]
These are the factors of the equation from this we can find the values of \[x\]
We will get x values as \[3\] and \[7\].
\[x=3,x=7\]
Note: We can also solve this by applying square root on both sides of the equation. By this we can simply get the solution. But we should be aware that we have to consider the square root as \[\pm \] then only we will get both the solutions otherwise we will get only one.
Complete step by step answer:
Given equation
\[{{\left( x-5 \right)}^{2}}=4\]
First we have to expand the LHS using the formula
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
By applying above formula we can expand our LHS as
\[{{\left( x-5 \right)}^{2}}={{x}^{2}}-2\times x\times 5+{{5}^{2}}\]
From this we will get
\[\Rightarrow {{x}^{2}}-10x+25\]
Now we have to substitute this in our equation. Our equation will look like
\[\Rightarrow {{x}^{2}}-10x+25=4\]
Now we will solve this quadratic equation using the factor method.
For this we have to make our RHS as \[0\] by transferring \[4\] to LHS. We will get
\[\Rightarrow {{x}^{2}}-10x+21=0\]
For solving using factor methods we have to split the middle term.
We have split the middle term in such a way that their sum is equal to the middle term and product is equal to first and last terms.
To get these numbers we have to find prime factorization of the product of first and last terms.
The product we will get from our equation is \[21\]
The prime factorization of \[21\] will be like
\[\Rightarrow 21=3\times 7\]
So we can see that \[-3\] and \[-5\] can satisfy our condition.
\[\begin{align}
& -3\times -7=21 \\
& -3+\left( -7 \right)=-10 \\
\end{align}\]
So we can split our middle term now. By splitting it we will get
\[\Rightarrow {{x}^{2}}-3x-7x+21=0\]
Now we have to take common terms .
From the equation we can take \[x\] as common from the first two terms and \[7\] as common from the next two terms.
Then the equation will look like
\[\Rightarrow x\left( x-3 \right)-7\left( x-3 \right)\]
\[\Rightarrow \left( x-3 \right)\left( x-7 \right)\]
These are the factors of the equation from this we can find the values of \[x\]
We will get x values as \[3\] and \[7\].
\[x=3,x=7\]
Note: We can also solve this by applying square root on both sides of the equation. By this we can simply get the solution. But we should be aware that we have to consider the square root as \[\pm \] then only we will get both the solutions otherwise we will get only one.
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